Can anyone please help me solve this one?
the result have to be in positive integers

Step-by-step explanation:

(a²-b²)(c²-d²) = a²c² - a²d² - b²c² + b²d²

(b²-d²)(c²-a²) = b²c² - a²b² - c²d² + a²d²

the sum of both is

a²c² + b²d² - a²b² - c²d² = a²(c²-b²) + d²(b²-c²) =

= a²(c²-b²) - d²(c²-b²) = (c²-b²)(a²-d²) = 2021

after a little check we find 2021 is only divisible by 43 and 47 (or 1 and 2021).

so, it is clear that our expression resembles

43×47 = 2021

that means that e.g. c²-b² = 43, and a²-d² = 47

or the other way around, it does not matter.

what squared integer numbers give a difference of 43 ? and 47 ?

squared numbers

1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256

289 324 361 400 441 484 529 576

aha ! 484 - 441 = 43

and 576 - 529 = 47

so, we get

a = sqrt(576) = 24

b = sqrt(441) = 21

c = sqrt(484) = 22

d = sqrt(529) = 23

one could also use the formula for consecutive squared numbers

(n+1)² = n² + 2n + 1

under the assumption that the 2 numbers are convective (which it turns out they are in this example) we get

(n+1)² - n² = 2n + 1 = 43

2n = 42

n = 21

n+1 = 22

and for 47

2n + 1 = 47

2n = 46

n = 23

n+1 = 24

u bout ugly as   a mf

step-by-step explanation: lol